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Washington Real Estate Fundamentals Lesson 18: Real Estate Math © 2011 Rockwell Publishing

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Solving Math Problems Four basic steps 1. Read the question. What answer are you being asked to find? 2. Write down the formula. 3. Substitute. Replace elements in formula with relevant numbers from problem. 4. Calculate the answer. © 2011 Rockwell Publishing

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Solving Math Problems Using formulas Each of these choices expresses the same formula, but in a way that lets you solve it for A, B, or C: A = B × C B = A ÷ C C = A ÷ B Choose the one that isolates the unknown. © 2011 Rockwell Publishing

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Solving Math Problems Isolating the unknown The unknown is the element that you’re trying to determine. The unknown should always sit alone on one side of the equals sign. All the information that you already know should be on the other side. © 2011 Rockwell Publishing

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Solving Math Problems Example If a lot has an area of 7,000 square feet and is 100 feet long, what is its width? Formula needed: Area = Length × Width Substitute numbers from problem: 7,000 sq. ft. = 100 ft. × Width Width is the unknown, so isolate Width by switching formula to Width = Area ÷ Length. Calculate: 7,000 sq. ft. ÷ 100 ft. = 70 ft. Width = 70 feet, so the lot is 70 feet wide. © 2011 Rockwell Publishing

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Decimal Numbers Converting fraction to decimal Calculators use only decimals, not fractions. If problem contains a fraction, convert it to a decimal: Divide top number (numerator) by bottom number (denominator). 1/4 = 1 ÷ 4 = /3 = 1 ÷ 3 = /8 = 5 ÷ 8 = © 2011 Rockwell Publishing

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Decimal Numbers Converting percentage to decimal If problem contains a percentage, may be necessary to: convert percentage to decimal do calculations to find answer then convert decimal back to percentage (Calculator’s percent key converts percentage to decimal automatically.) © 2011 Rockwell Publishing

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Decimal Numbers Converting percentage to decimal To convert a percentage to a decimal: Remove percent sign. Move decimal point two places to the left. (May be necessary to add a zero.) 2% becomes % becomes % becomes % becomes 1.23 © 2011 Rockwell Publishing

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Decimal Numbers Converting decimal to percentage To convert a decimal to a percentage, reverse the process: Move decimal point two places to the right. Add percent sign..02 = 2%.175 = 17.5%.80 = 80% 1.23 = 123% © 2011 Rockwell Publishing

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Summary Solving Math Problems Four steps: read problem, write formula, substitute, calculate Using formulas Isolating the unknown Converting fraction to decimal number Converting percentage to decimal number Converting decimal number to percentage © 2011 Rockwell Publishing

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Area Problems Real estate agent may need to calculate area of lot, building, or room. Area usually stated in square feet or square yards. Formula needed depends on shape. Area problem may also involve calculating other elements, such as cost per square foot or rental rate. © 2011 Rockwell Publishing

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Area Problems Rectangle formula: A = L × W Formula for calculating area of square or rectangular space: Area = Length × Width A = L × W Area Width Length

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Rectangles Example An office is 27 feet wide by 40 feet long. It rents for $2 per square foot per month. How much is the monthly rent? Step 1: Calculate area A = 27 ft. × 40 ft. A = 1,080 sq. ft. Step 2: Calculate rent Rent = 1,080 sq. ft. × $2 per sq. ft. Rent = $2,160 per month © 2011 Rockwell Publishing

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Rectangles Square yards Some problems express area in square yards rather than square feet. Remember: 1 square yard = 9 square feet 1 yard = 3 feet 1 square yard is 3 feet on each side 3 feet × 3 feet = 9 square feet Divide square footage by 9 to find number of square yards. © 2011 Rockwell Publishing

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Area Problems Triangle formula: A = ½ B × H Formula for calculating area of triangle: Area = ½ Base × Height A = ½ B × H

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Triangles Order of operations Note that in applying triangle formula, order of operations doesn’t matter. These will all reach same result: A = ½ B × H A = B × ½ H A = (B × H) ÷ 2 © 2011 Rockwell Publishing

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Triangles Example A triangular lot is 140 feet long and 50 feet wide at its base. What is the area? A = (½ × 50 ft.) × 140 ft. A = 25 ft. × 140 ft. A = 3,500 sq. ft.

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Triangles Example, continued A triangular lot is 140 feet long and 50 feet wide at its base. What is the area? A = (½ × 50 ft.) × 140 ft. A = 25 ft. × 140 ft. A = 3,500 sq. ft. Same result from this: A = 50 ft. × (½ × 140 ft.) A = 50 ft. × 70 ft. A = 3,500 sq. ft. And from this: A = (50 ft. × 140 ft.) ÷ 2 A = 7,000 sq. ft. ÷ 2 A = 3,500 sq. ft.

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Area Problems Odd shapes To find area of an irregular shape: Divide figure up into squares, rectangles, and triangles. Find area of each of the shapes that make up the figure. Add the areas together. © 2011 Rockwell Publishing

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Odd Shapes Example The lot’s western side is 60 feet long. Its northern side is 100 feet long, but its southern side is 120 feet long. To find the area of this lot, break it into a rectangle and a triangle.

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Odd Shapes Example, continued Area of rectangle: A = 60 ft. × 100 ft. A = 6,000 sq. feet

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Odd Shapes Example, continued To find length of triangle’s base, subtract length of northern boundary from length of southern boundary. 120 ft. – 100 ft. = 20 ft. Area of triangle: A = (½ × 20 ft.) × 60 ft. A = 600 sq. ft.

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Odd Shapes Example, continued Total area of odd shape: 6,000 sq. ft. (rectangle) sq. ft. (triangle) 6,600 sq. ft.

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Odd Shapes Avoid counting same section twice Common mistake when working with odd shapes: Calculating area of part of the figure twice. Can happen with a figure like this one.

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Odd Shapes Example Wrong way to calculate area of this lot: 25 × 50 = 1,250 sq. ft. 40 × 20 = 800 sq. ft. 1, = 2,050 sq. ft. This counts middle of shape (green area) twice.

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Odd Shapes Example, continued Avoid the mistake by breaking the shape down like this instead. Find height of smaller rectangle by subtracting height of top rectangle (25 ft.) from height of whole shape (40 ft.). 40 – 25 = 15 ft.

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Odd Shapes Example, continued Now calculate area of each rectangle and add them together: 25 × 50 = 1,250 sq. ft. 20 × 15 = 300 sq. ft. 1, = 1,550 sq. ft.

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Odd Shapes Example, continued Another way to break the odd shape down into rectangles correctly: To find width of rectangle on right, subtract width of left rectangle from width of whole shape: 50 – 20 = 30 feet

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Odd Shapes Example, continued Now calculate area of each rectangle and add them together: 40 × 20 = 800 sq. ft. 30 × 25 = 750 sq. ft = 1,550 sq. ft.

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Odd Shapes Narrative problems Some odd shape problems are expressed only in narrative form, without a diagram. Useful to draw the shape yourself, then break it down into rectangles and triangles. © 2011 Rockwell Publishing

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Odd Shapes Example A lot’s boundary begins at a certain point and runs due south for 319 feet, then east for 426 feet, then north for 47 feet, and then back to the point of beginning. To solve this problem, first draw the shape.

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Odd Shapes Example, continued Break it down into a rectangle and a triangle as shown. Subtract 47 from 319 to find height of triangular portion: 319 – 47 = 272 ft.

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Odd Shapes Example, continued Calculate area of rectangle: 426 × 47 = 20,022 sq. ft.

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Odd Shapes Example, continued Calculate area of triangle: (½ × 426) × 272 = 57,936 sq. ft.

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Odd Shapes Example, continued Add together area of rectangle and area of triangle to find the lot’s total square footage: 20, ,936 = 77,958

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Volume Problems Area: Measurement of a two-dimensional space. Volume: Measurement of a three-dimensional space. Width, length, and height. Cubic feet or cubic yards instead of square feet or square yards. © 2011 Rockwell Publishing

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Volume Problems Formula: V = L × W × H To calculate volume, use this formula: Volume = Length × Width × Height V = L × W × H © 2011 Rockwell Publishing

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Volume Problems Cubic yards If problem asks for cubic yards: 1 cubic yard = 27 cubic feet 1 cubic yard measures 3 feet on each dimension 3 feet × 3 feet × 3 feet = 27 cubic feet Divide cubic footage by 27 to find number of cubic yards. © 2011 Rockwell Publishing

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Volume Problems Example A trailer is 40 feet long, 9 feet wide, and 7 feet high. How many cubic yards does it contain? 40 × 9 × 7 = 2,520 cubic feet 2,520 ÷ 27 = cubic yards © 2011 Rockwell Publishing

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Summary Area and Volume Problems Area of square or rectangle: A = L × W 1 square yard = 9 square feet Area of triangle: A = ½ B × H Divide odd shapes into squares, rectangles, and triangles. Volume: V = L × W × H 1 cubic yard = 27 cubic feet © 2011 Rockwell Publishing

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Percentage Problems Many math problems ask you to find a certain percentage of another number. Example: What is 85% of $150,000? This means you need to multiply that other number by the percentage: $150,000 ×.85 = $127,500 (Note conversion of percentage to decimal, explained earlier.) © 2011 Rockwell Publishing

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Percentage Problems Formula: P = W × % Basic formula for solving percentage problems: Part = Whole × Percentage P = W × % © 2011 Rockwell Publishing

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Percentage Problems Formula: P = W × % “Whole” is larger figure. Example: property’s sales price “Part” is smaller figure. Example: amount of commission owed A percentage of the whole equals the part. © 2011 Rockwell Publishing

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Percentage Problems Isolating the unknown Use basic formula (P = W × %) if part is the unknown. If whole is the unknown: W = P ÷ % If percentage is the unknown: % = P ÷ W © 2011 Rockwell Publishing

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Percentage Problems Multiply or divide? Knowing whether to divide or multiply can be hardest part of solving percentage problems. If missing element is the part (smaller number), it’s a multiplication problem. If missing element is either the whole (larger number) or the percentage, it’s a division problem. © 2011 Rockwell Publishing

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Percentage Problems Types of problems Many real estate math problems are basically percentage problems: commission problems loan problems (interest and principal) profit or loss problems capitalization problems Percentage sometimes referred to as “rate.” Examples: 7% commission rate, 5% interest rate, 10% rate of return © 2011 Rockwell Publishing

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Commission Problems For commission problems, use P = W × %. In commission problems: Percentage (%) is commission rate. Whole (W) is amount commission based on. Usually sales price of house. Part (P) is amount of commission. © 2011 Rockwell Publishing

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A home sells for $300,000. Listing firm is paid a 6% commission on the sales price. Listing broker is entitled to 60% of the commission. How much is the broker’s share? Step 1: Calculate total commission (6%) P = W × % P = $300,000 ×.06 P = $18,000 Step 2: Calculate broker’s share (60%) P = $18,000 ×.60 P = $10,800 Commission Problems Example

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Commission Problems Two-tiered commission Some problems involve two-tiered commission: one rate on sales price up to certain amount different rate on amount by which price exceeds that amount © 2011 Rockwell Publishing

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Listing firm’s commission is 7% on first $300,000 of home’s sales price, 5% on any amount over $300,000. If commission is $23,500, how much was the sales price? Step 1: Calculate commission on first $300,000 P = W × % P = $300,000 ×.07 P = $21,000 Step 2: Subtract to find commission on rest of price $23,500 – $21,000 = $2,500 Two-tiered Commission Example

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© 2011 Rockwell Publishing Step 3: Calculate amount of price over $300,000 P = W × % $2,500 = W ×.05 Isolate unknown: W = $2,500 ÷.05 W = $50,000 Step 4: Add to find total sales price $300,000 + $50,000 = $350,000 Two-tiered Commission Example, continued

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Summary Percentage Problems Percentage formula: P = W × % W = P ÷ % % = P ÷ W Types of percentage problems: commissions, loans, profit or loss, capitalization Commission problems: –% is commission rate –W is usually sales price of home –P is commission amount © 2011 Rockwell Publishing

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Loan Problems Loan problems include: interest problems principal balance problems For loan problems, again use P = W × %. Now P is amount of interest. W is loan amount or principal balance. % is interest rate, expressed as an annual rate. © 2011 Rockwell Publishing

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Loan Problems Interest problems Some problems state interest amount (the part, P) in semiannual, quarterly, or monthly installments. Semiannual: Twice a year (two payments per year). Quarterly: Four times a year (four payments per year). First step in problems like these: convert interest payments into annual figure. © 2011 Rockwell Publishing

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A lender makes an interest-only loan of $140,000. The interest rate is 6.5%. How much does the annual interest come to? P = W × % P = $140,000 ×.065 P = $9,100 Interest Problems Example: Finding annual interest

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© 2011 Rockwell Publishing The interest portion of a loan’s monthly payment is $ The loan balance is $92,000. What is the interest rate? Step 1: Convert monthly to annual figure $ × 12 = $6,210 annual interest Step 2: Calculate interest rate P = W × % Substitute: $6,210 = $92,000 × % Isolate the unknown: % = $6,210 ÷ $92,000 % =.0675, or 6.75% Interest Problems Example: Finding interest rate

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Loan Problems Principal balance problems Some problems give interest amount and interest rate and ask you to find loan amount or principal balance. Once again, use basic percentage formula, P = W × % © 2011 Rockwell Publishing

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A real estate loan calls for semiannual interest-only payments of $3,250. The interest rate is 9%. What is the loan amount? Step 1: Convert semiannual to annual figure $3,250 × 2 = $6,500 annual interest Step 2: Calculate loan amount P = W × % Substitute: $6,500 = W ×.09 Isolate the unknown: W = $6,500 ÷.09 W = $72, Principal Balance Problems Example: Finding loan amount

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© 2011 Rockwell Publishing The interest portion of a loan’s tenth monthly payment is $ The interest rate is 7%. What is the loan balance prior to this payment? Step 1: Convert monthly to annual figure $ × 12 = $3, (annual interest) Step 2: Calculate principal balance P = W × % Substitute: $3,080 = W ×.07 Isolate the unknown: W = $3,080 ÷.07 W = $44,000 Principal Balance Problems Example: Finding current balance

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Principal Balance Problems Principal and interest payments Some problems give the monthly principal and interest payment (instead of just the interest portion of monthly payment) and ask how much payment reduces balance. These require an additional step: subtracting interest portion from payment amount. © 2011 Rockwell Publishing

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A loan’s balance is $96,000. The interest rate is 8%. The monthly principal and interest payment is $ How much will this payment reduce the loan balance? Step 1: Calculate annual interest P = W × % $96,000 ×.08 = $7,680 annual interest Step 2: Divide to find interest portion of payment $7,680 ÷ 12 = $640 Step 3: Subtract to find principal portion $ – $ = $64.41 Principal Balance Problems Example: Finding principal reduction

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Principal Balance Problems Multiple P&I payments If question asks how much loan balance will be after second, third, or sixth principal and interest payment: calculate first payment’s effect on principal balance repeat steps again as many times as necessary, using new balance each time © 2011 Rockwell Publishing

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Continuing with previous example ($96,000 8%, $ P&I pyt, first pyt reduces balance by $64.41): $96,000 – $64.41 = $95, (balance after Pyt 1) $95, ×.08 = $7, (annual interest) $7, ÷ 12 = $ (interest portion of Pyt 2) $ – $ = $64.84 (principal portion of Pyt 2) $95, – $64.84 = $95, (balance after Pyt 2) So second payment reduces loan balance to $95, For balance after third payment, repeat these steps. Principal Balance Problems Example: Multiple P&I payments

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Summary Loan Problems Use percentage formula: P = W × % –P is interest amount –% is interest rate –W is loan amount or principal balance Convert semiannual, quarterly, or monthly interest into annual interest. © 2011 Rockwell Publishing

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Profit or Loss Problems Another common type of percentage problem: calculating property owner’s profit or loss. Use this variation on percentage formula: Now = Then × % “Then” is property’s value at earlier point. “Now” is property’s value at later point. Percentage refers to rate of profit (appreciation) or loss (depreciation). © 2011 Rockwell Publishing

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Profit or Loss Problems Then and Now formula Basic “Then and Now” formula: Now = Then × % To isolate the unknown, change it to: Then = Now ÷ % or % = Now ÷ Then © 2011 Rockwell Publishing

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Profit or Loss Problems Percentage to use in formula Key to solving these problems: finding correct percentage to use in formula. Percentage is 100% plus the percentage of profit, or minus the percentage of loss. If no profit or loss, Now value is 100% of Then value. If profit, Now value is greater than 100% of Then value. If loss, Now value less than 100% of Then. © 2011 Rockwell Publishing

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Percentage to Use in Formula Property sold at a loss For loss, subtract percentage of loss from 100% to find percentage to use in formula. Example: 30% loss House didn’t sell for 30% of original value. It sold for 30% less than original value. 100% – 30% = 70% So Now value is 70% of Then value. Use 70% (.70) in formula. © 2011 Rockwell Publishing

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A seller sells her house for $420,000, which represents a 30% loss. How much did she originally pay for it? Step 1: Determine percentage to use in formula 100% – 30% = 70%, or.70 Step 2: Calculate Then value Now = Then × % Substitute: $420,000 = Then ×.70 Isolate the unknown: Then = $420,000 ÷.70 Then = $600,000 Percentage to Use in Formula Example: Property sold at a loss

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Percentage to Use in Formula Property sold for a profit If property sold for a profit, add percentage of profit to 100% to find percentage to use in formula. Example: 30% profit Means house sold for 130% of what seller originally paid. 100% + 30% = 130% So Now value is 130% of Then value. Use 130% (1.30) in formula. © 2011 Rockwell Publishing

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Percentage to Use in Formula Property sold for a profit Note that if seller sells house for 130% of what she paid for it, that doesn’t means she has made a 130% profit. She received 100% of what she paid, plus 30%. She made a 30% profit. © 2011 Rockwell Publishing

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A seller sells her house for $420,000, which represents a 30% profit. How much did she originally pay for it? Step 1: Determine percentage to use in formula 100% + 30% = 130%, or 1.30 Step 2: Calculate Then value Now = Then × % Substitute: $420,000 = Then × 1.30 Isolate the unknown: Then = $420,000 ÷ 1.30 Then = $323,077 Percentage to Use in Formula Example: Property sold for a profit

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Profit or Loss Problems Calculating selling price Examples so far gave selling price and profit or loss percentage, asked how much seller originally paid for property (the Then value). Other problems will: give original price and percentage of profit or loss ask how much property is being sold for (the Now value) © 2011 Rockwell Publishing

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A seller bought his house four years ago for $200,000 and sold it this year for 5% more than he paid for it. How much did he sell it for? Step 1: Determine percentage to use in formula 100% + 5% = 105%, or 1.05 Step 2: Calculate Now value Now = Then × % Substitute: Now = $200,000 × 1.05 Now = $210,000 Profit or Loss Problems Example: Calculating selling price

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Profit or Loss Problems Calculating profit or loss percentage Other problems give original price and selling price, ask for percentage of profit or loss. © 2011 Rockwell Publishing

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A seller bought his house four years ago for $200,000 and sold it this year for $210,000. What was his percentage of profit on the sale? Now = Then × % Substitute: $210,000 = $200,000 × % Isolate the unknown: % = $210,000 ÷ $200,000 % = 1.05, or 105% 105% – 100% = 5% The seller made a 5% profit on the sale. Profit or Loss Problems Example: Calculating profit

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© 2011 Rockwell Publishing Calculating a loss works the same way: A seller bought his house seven years ago for $184,000 and sold it this year for $160,000. What was his percentage of loss on the sale? Now = Then × % Substitute: $160,000 = $184,000 × % Isolate the unknown: % = $160,000 ÷ $184,000 % =.869, or 87% 100% – 87% = 13% The seller took a 13% loss on the sale. Profit or Loss Problems Example: Calculating loss

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Profit or Loss Problems Appreciation or depreciation Profit or loss problem may also be expressed in terms of appreciation or depreciation. Problem solved same way as ordinary profit or loss problem. If problem asks how much property appreciated or depreciated per year over several years, repeat same calculation for each year. © 2011 Rockwell Publishing

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A property is currently worth $420,000. It has depreciated four and a half percent per year for the past five years. What was the property worth five years ago? Step 1: Find percentage to use in formula 100% – 4.5% = 95.5%, or.955 Step 2: Calculate value one year ago Now = Then × % Substitute: $420,000 = Then ×.955 Isolate the unknown: Then = $420,000 ÷.955 Then = $439,791 (value 1 year ago) Profit or Loss Problems Example: Compound depreciation

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© 2011 Rockwell Publishing Now that you know how much the house was worth one year ago, calculate the Then value four more times, to find how much the house was worth five years ago: $439,791 ÷.955 = $460,514 (value 2 years ago) $460,514 ÷.955 = $482,214 (value 3 years ago) $482,214 ÷.955 = $504,936 (value 4 years ago) $504,936 ÷.955 = $528,729 (value 5 years ago) The house was worth around $528,729 five years ago. Profit or Loss Problems Example, continued

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Profit or Loss Problems Compound appreciation If you’re told that a property gained value at a particular rate over several years, you’ll use the same process. The difference is that you’ll need to add the rate of change to 100%, instead of subtracting it from 100%. © 2011 Rockwell Publishing

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A property is currently worth $380,000. It has appreciated in value 4% per year for the last four years. What was it worth four years ago? Step 1: Find percentage to use in formula 100% + 4% = 104%, or 1.04 Step 2: Calculate Then value for four years $380,000 ÷ 1.04 = $365,385 (value 1 year ago) $365,385 ÷ 1.04 = $351,332 (value 2 years ago) $351,332 ÷ 1.04 = $337,819 (value 3 years ago) $337,819 ÷ 1.04 = $324,826 (value 4 years ago) Profit or Loss Problems Example: Compound appreciation

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Summary Profit or Loss Problems Now = Then × Percentage To find percentage to use in formula: –If loss or depreciation, subtract rate of change from 100%. –If profit or appreciation, add rate of change to 100%. Compound appreciation and depreciation: repeat profit or loss calculation as needed. © 2011 Rockwell Publishing

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Capitalization Problems Capitalization: In appraisal of income-producing property, process used to convert property’s income into estimate of its value. Value is price an investor would be willing to pay for the property. Property’s annual net income is investor’s return on the investment (ROI). © 2011 Rockwell Publishing

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Capitalization Problems Formula: I = V × % Capitalization problems are another type of percentage problem. Formula is another variation on P = W × % Part is property’s income, Whole is property’s value, % is capitalization rate. Income = Value × Capitalization Rate Value = Income ÷ Rate Rate = Income ÷ Value © 2011 Rockwell Publishing

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Capitalization Problems Capitalization rate Capitalization rate represents rate of return an investor would be likely to want on this investment. Investor who wants higher rate of return won’t pay as much for the property as investor who’s willing to accept lower rate of return. Higher cap rate means lower value. © 2011 Rockwell Publishing

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A property generates an annual net income of $48,000. An investor wants a 12% rate of return on his investment. How much could he pay for the property and realize his desired rate of return? Income = Value × Rate Substitute: $48,000 = Value ×.12 Isolate the unknown: Value = $48,000 ÷.12 Value = $400,000 The investor could pay $400,000 for the property and realize his desired 12% return. Thus, the property is worth $400,000 to this investor. Capitalization Problems Example: Calculating value

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© 2011 Rockwell Publishing An investment property is valued at $425,000 and its net income is $40,375. What is the capitalization rate? Income = Value × Rate Substitute: $40,375 = $425,000 × Rate Isolate the unknown: Rate = $40,375 ÷ $425,000 Rate =.095, or 9.5% Capitalization Problems Example: Finding the cap rate

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Capitalization Problems Changing the cap rate Capitalization rate is up to the investor, and depends on many factors. One investor might be satisfied with 9% return. Another might want 10.5% return from same property. Some problems ask how property’s value will change if different cap rate is applied. © 2011 Rockwell Publishing

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Using a 10% cap rate, a property is valued at $1,500,000. What would its value be using an 11.5% cap rate? Step 1: Calculate property’s annual net income Income = Value × Rate Substitute: Income = $1,500,000 ×.10 Income = $150,000 Capitalization Problems Example: Changing the cap rate

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© 2011 Rockwell Publishing Step 2: Calculate value at higher cap rate Income = Value × Rate Substitute: $150,000 = Value ×.115 Isolate the unknown: Value = $150,000 ÷.115 Value = $1,304,348 ($195,652 less) Capitalization Problems Example, continued

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© 2011 Rockwell Publishing A property with net income of $16,625 is valued at $190,000. If the cap rate is increased by 1%, what is the property’s value? Step 1: Find current capitalization rate Income = Value × Capitalization Rate Substitute: $16,625 = $190,000 × Rate Isolate the unknown: Rate = $16,625 ÷ $190,000 Rate =.0875 Capitalization Problems Example: Changing the cap rate

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© 2011 Rockwell Publishing Step 2: Increase cap rate by 1% 8.75% + 1% = 9.75%, or.0975 Step 3: Calculate new value Income = Value × Capitalization Rate Substitute: $16,625 = Value ×.0975 Isolate the unknown: Value = $16,625 ÷.0975 Value = $170,513 Capitalization Problems Example, continued

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Capitalization Problems Calculating net income Instead of property’s annual net income, some problems give annual gross income and: bad debt/vacancy factor list of operating expenses These must be subtracted from gross income to arrive at net income before capitalization formula can be applied. © 2011 Rockwell Publishing

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A six-unit apartment building rents three units for $650 a month and three units for $550 a month. Allow 5% for vacancies and uncollected rent. The annual operating expenses are $4,800 for utilities, $8,200 for property taxes, $1,710 for insurance, $5,360 for maintenance, and $2,600 for management fees. If the cap rate is 8¾%, what is the property’s value? Capitalization Problems Example: Calculating net income

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© 2011 Rockwell Publishing Step 1: Calculate gross income $550 × 3 × 12 = $19,800 (three units at $550/month) $650 × 3 × 12 = $23,400 (three units at $650/month) $19,800 + $23,400 = $43,200 (gross income) Step 2: Subtract 5% bad debt/vacancy factor $43,200 ×.05 = $2,160 $43,200 – $2,160 = $41,040 (effective gross income) Capitalization Problems Example, continued

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© 2011 Rockwell Publishing Step 3: Subtract operating expenses List of operating expenses adds up to $22,670 $41,040 – $22,670 = $18,370 (net income) Step 4: Calculate value with 8¾% cap rate Income = Value × Rate Substitute: $18,370 = Value ×.0875 Isolate the unknown: Value = $18,370 ÷.0875 Value = $209,943 Capitalization Problems Example, continued

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Capitalization Problems Calculating net income with OER Some problems give property’s operating expense ratio (OER) instead of list of operating expenses. OER is percentage of gross income that goes to pay operating expenses. Multiply gross income by OER to determine annual operating expenses. Then subtract expenses from gross income to determine net income. © 2011 Rockwell Publishing

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A store grosses $758,000 annually. It has an operating expense ratio of 87%. With a capitalization rate of 9¼%, what is its value? Step 1: Multiply gross income by OER $758,000 ×.87 = $659,460 (operating expenses) Step 2: Subtract expenses from gross income $758,000 – $659,460 = $98,540 (net income) Step 3: Apply capitalization formula (I = V × %) Value = $98,540 ÷.0925 Value = $1,065,297 Capitalization Problems Example: OER

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Summary Capitalization Problems Income = Value × Rate (I = V × %) –Income is annual net income –Rate is capitalization rate Higher the cap rate, the lower the value. Gross income Bad debt/vacancy factor Operating expenses OER: Operating expense ratio © 2011 Rockwell Publishing

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Tax Assessment Problems Some tax assessment problems can be solved using this variation on basic percentage formula (Part = Whole × %): Tax = Assessed Value × Tax Rate © 2011 Rockwell Publishing

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Tax Assessment Problems Assessment ratio Problem may state property’s assessed value. Or it may give market value and assessment ratio, and assessed value must be calculated. Multiply market value by assessment ratio to find assessed value. Example: Property’s market value is $500,000 and assessment ratio is 80%. $500,000 ×.80 = $400,000 © 2011 Rockwell Publishing

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The property’s market value is $410,000. It’s subject to a 25% assessment ratio and an annual tax rate of 2.5%. How much is the annual tax the owner must pay? Step 1: Calculate assessed value $410,000 ×.25 = $102,500 Step 2: Calculate tax Tax = Assessed Value × Tax Rate Substitute: Tax = $102,500 ×.025 Tax = $2, The property owner must pay $2, Tax Assessment Problems Example: Assessment ratio

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Tax Assessment Problems Tax rate per $100 or $1,000 In some problems, tax rate not expressed as a percentage. Instead, stated as a dollar amount per hundred dollars or per thousand dollars of assessed value. Divide value by 100 or 1,000 to find number of $100 or $1,000 increments. Then multiply that number by tax rate. © 2011 Rockwell Publishing

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A property is assessed at $125,000. The tax rate is $2.10 per hundred dollars of assessed value. What is the annual tax? Step 1: Find number of increments in assessed value $125,000 ÷ 100 = 1,250 ($100 increments) Step 2: Multiply number of increments by tax rate 1,250 × $2.10 = $2,625 (annual tax) Tax Assessment Problems Example: Tax rate per $100

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© 2011 Rockwell Publishing A property is assessed at $396,000. The tax rate is $14.25 per thousand dollars of assessed value. What is the annual tax? Step 1: Find number of increments in assessed value $396,000 ÷ 1,000 = 396 ($1,000 increments) Step 2: Multiply number of increments by tax rate 396 × $14.25 = $5,643 (annual tax) Tax Assessment Problems Example: Tax rate per $1,000

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Tax Assessment Problems Tax rate in mills One other way in which a tax rate may be expressed: mills per dollar of assessed value. One mill is one-tenth of one cent (.001), or one-thousandth of a dollar. To convert mills to a percentage rate in decimal form, divide by 1,000. Example: 30 mills ÷ 1,000 =.03 (3%) © 2011 Rockwell Publishing

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A property is assessed at $290,000 and the tax rate is 23 mills per dollar of assessed value. How much is the annual tax? Step 1: Convert mills to a percentage rate 23 mills/dollar ÷ 1,000 =.023 (2.3%) Step 2: Multiply assessed value by tax rate Tax = Assessed Value × Tax Rate Substitute: Tax = $290,000 ×.023 Tax = $6,670 Tax Assessment Problems Example: Tax rate in mills

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Summary Tax Assessment Problems Tax = Assessed Value × Tax Rate Assessed value Assessment ratio Dollar amount per $100 or $1,000 of value Mills © 2011 Rockwell Publishing

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Seller’s Net Problems These ask how much seller needs to sell property for in order to get specified net amount from sale. © 2011 Rockwell Publishing

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Seller’s Net Problems Steps for finding minimum price 1.Add together: seller’s desired net costs of sale, except for commission 2.Subtract commission rate from 100%. 3.Divide result of Step 1 by result of Step 2. If problem states mortgage balance to be paid off, treat that as one of the costs of sale. © 2011 Rockwell Publishing

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A seller wants to net $120,000 from the sale of his property. He’ll pay $1,650 in attorney’s fees, $700 for the escrow fee, $550 for repairs, and a 6% commission. What’s the minimum he can sell the property for? Step 1: Add desired net and costs of sale $120,000 + $1,650 + $700 + $550 = $122,900 Step 2: Subtract commission rate from 100% 100% - 6% = 94%, or.94 Step 3: Calculate minimum sales price $122,900 ÷.94 = $130, Seller’s Net Problems Example

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© 2011 Rockwell Publishing A seller wants to net $24,000 from selling her home. She will have to pay $3,500 in closing costs, $1,600 in discount points, and a 7% commission. She’ll also have to pay off the mortgage balance, which is $86,050. What is the minimum amount she’ll have to sell the home for? Step 1: Add desired net, mortgage, and sale costs $24,000 + $86,050 + $3,500 + $1,600 = $115,150 Seller’s Net Problems Example

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© 2011 Rockwell Publishing Step 2: Subtract commission rate from 100% 100% – 7% = 93%, or.93 Step 3: Calculate minimum sales price $115,150 ÷.93 = $123, Seller’s Net Problems Example, continued

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Summary Seller’s Net Problems Desired Net + Costs of Sale + Loan Payoff Subtract commission rate from 100% Divide Step 1 total by Step 2 rate. Result is how much property must sell for. © 2011 Rockwell Publishing

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Proration Problems Proration: Dividing an expense proportionally, when one party is responsible for only part of the expense. Items often prorated in real estate transactions include: property taxes insurance premiums rent mortgage interest © 2011 Rockwell Publishing

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Proration Problems Closing date is proration date In most cases, property expenses prorated as of closing date. Seller’s responsibility for expense usually ends on closing date. Buyer’s responsibility for expense usually begins on closing date. © 2011 Rockwell Publishing

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Proration Problems 3 Steps 1. Calculate per diem (daily) rate of expense. 2. Determine number of days one party is responsible for expense. 3. Multiply per diem rate by number of days. Share = Rate × Days © 2011 Rockwell Publishing

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Proration Problems 365 days or 360 days Problem will say whether to use: 365-day year (calendar year), or 360-day year (banker’s year). With 365-day year, use exact number of days in each month. In 360-day year, each month has 30 days. © 2011 Rockwell Publishing

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Proration Problems Property taxes In Washington: General real estate taxes levied annually. Tax year: January 1 – December 31 First installment due April 30 (covers January through June) Second installment due October 31 (covers July through December) © 2011 Rockwell Publishing

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Property Tax Prorations Paid in advance or in arrears Seller may already have paid tax installment or full year’s taxes. Buyer owes seller prorated share at closing. Debit for buyer, credit for seller. Or seller may not yet have paid taxes. Seller owes buyer prorated share at closing. Debit for seller, credit for buyer. Buyer’s responsibility starts on closing date. © 2011 Rockwell Publishing

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The annual property taxes are $2,860. The seller has paid the first installment, but not the second. The sale is closing on August 10. How much will the seller have to pay in taxes at closing? Use a 360-day year. Step 1: Calculate per diem rate $2,860 ÷ 360 = $7.94 Step 2: Count number of days (starting July 1) 30 (July) + 9 (August) = 39 days Step 3: Multiply rate by number of days $7.94 × 39 = $ (seller’s debit, buyer’s credit) Property Tax Prorations Example: Taxes in arrears

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© 2011 Rockwell Publishing A home sale is due to close on September 20. The annual property taxes, $4,924, have already been paid through the end of the year. How much will the buyer owe at closing for property taxes? Use a 365-day year. Step 1: Calculate per diem rate $4,924 ÷ 365 = $13.49 Step 2: Count number of days 11 (Sept) + 31 (Oct) + 30 (Nov) + 31 (Dec) = 103 days Step 3: Multiply per diem rate by number of days $13.49 × 103 = $1, (buyer’s debit, seller’s credit) Property Tax Prorations Example: Taxes paid in advance

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Proration Problems Hazard insurance Property insurance is paid for in advance. Seller entitled to refund of prorated share of insurance premium for coverage extending beyond closing date. Refund is credit for seller. Not a debit for buyer, unless buyer is assuming policy. © 2011 Rockwell Publishing

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The annual premium for the sellers’ hazard insurance policy is $1,350. It’s been paid for through next March, but the sale is closing November 12. The sellers are responsible for insuring the property until the day after closing. How much is their refund? Use a 360-day year. Step 1: Calculate per diem rate $1,350 ÷ 360 = $3.75 Insurance Prorations Example

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© 2011 Rockwell Publishing Step 2: Count number of days 18 (Nov.) (December to March) = 138 days Step 3: Multiply per diem rate by number of days $3.75 × 138 = $ (sellers’ credit) Insurance Prorations Example, continued

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Proration Problems Rent When rental property sold, seller owes buyer prorated share of any rent paid in advance. Tenants generally expected to pay rent at beginning of month. © 2011 Rockwell Publishing

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The sale of an apartment house is closing January 12. The seller collected January rent from all tenants; it totals $14,400. How much will the seller owe the buyer for rent at closing? Use the exact number of days in the month. Step 1: Calculate per diem rate $14,400 ÷ 31 days = $ per diem Rent Prorations Example

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© 2011 Rockwell Publishing Step 2: Count number of days January 12 through January 31 = 20 days Step 3: Multiply per diem rate by number of days $ × 20 = $9, (seller’s debit, buyer’s credit) Rent Prorations Example, continued

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Proration Problems Mortgage interest For interest prorations, remember that mortgage interest is almost always paid: on a monthly basis in arrears (at end of month in which it accrues) If problem doesn’t give amount of annual interest, first use loan amount and interest rate to calculate it. Then do the other proration steps. © 2011 Rockwell Publishing

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Mortgage Interest Prorations Two types of prorated interest Two types of mortgage interest usually have to be prorated as of closing date and paid at closing: seller’s final interest payment buyer’s prepaid interest (interim interest) For both types, lender charges interest for day of closing. © 2011 Rockwell Publishing

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Ann is selling her home for $275,000, closing May 14. Her mortgage at 7% interest has a balance of $212,500. How much is her final interest payment? (Use 365 days.) Step 1: Calculate annual interest $212,500 ×.07 = $14,875 Step 2: Calculate per diem rate $14,875 ÷ 365 = $40.75 Mortgage Interest Prorations Example: Seller’s final payment

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© 2011 Rockwell Publishing Step 3: Count number of days May 1 through May 14 = 14 days Step 4: Multiply per diem by number of days $40.75 × 14 = $ Mortgage Interest Prorations Example, continued

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Mortgage Interest Prorations Buyer’s prepaid interest Prepaid interest: At closing, buyer is charged interest for closing date through end of month in which closing occurs. Example: Sale closing on April 8 Buyer’s first loan payment due June 1. June 1 payment includes May interest, but not April interest. At closing, buyer must pay interest for April 8 through April 30. © 2011 Rockwell Publishing

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A home buyer has a $350,000 loan at 5.5% interest. The closing date is January 17. How much prepaid interest will the buyer have to pay? (Use a 360-day year.) Step 1: Calculate annual interest $350,000 ×.055 = $19,250 Step 2: Calculate per diem rate $19,250 ÷ 360 = $53.47 Mortgage Interest Prorations Example: Buyer’s prepaid interest

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© 2011 Rockwell Publishing Step 3: Count number of days January 17 through January 30 = 14 days Step 4: Multiply per diem rate by number of days $53.47 × 14 = $ Mortgage Interest Prorations Example, continued

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Summary Proration Problems Closing date 365-day year vs. 360-day year Payment in advance or in arrears Steps: –Calculate per diem rate –Count number of days –Multiply per diem rate by number of days © 2011 Rockwell Publishing

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